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CAT 2018 Slot 2 — QA Question 3

MeansEasy

The arithmetic mean of x, y and z is 80, and that of x, y, z, u and v is 75, where u = (x + y)/2 and v = (y + z)/2. If x ≥ z, then the minimum possible value of x is

Answer & solution

Answer: 105

Solution

Easy

Turn both means into sum equations. Subtracting them yields u+vu+v, and since u,vu,v are expressions in x,y,zx,y,z this produces a third linear relation. Solve to pin down yy and the sum x+zx+z, then use x≥zx\ge z to minimise xx.

1

Write the two means as sums.

x+y+z=80×3=240(1)x+y+z+u+v=75×5=375(2)\begin{aligned} &x+y+z = 80\times 3 = 240 \quad\text{(1)}\\ &x+y+z+u+v = 75\times 5 = 375 \quad\text{(2)} \end{aligned}
2

Subtract to get u+vu+v, then expand. From (2)−(1)(2)-(1) we get u+v=135u+v=135. Substitute u=x+y2, v=y+z2u=\dfrac{x+y}{2},\ v=\dfrac{y+z}{2}:

x+y2+y+z2=135⇒ x+2y+z=270(3)\begin{aligned} &\frac{x+y}{2}+\frac{y+z}{2} = 135\\ &\Rightarrow\ x+2y+z = 270 \quad\text{(3)} \end{aligned}
3

Eliminate to find yy and x+zx+z. Subtract (1)(1) from (3)(3):

(x+2y+z)−(x+y+z)=270−240⇒ y=30⇒ x+z=240−30=210(from (1))\begin{aligned} &(x+2y+z)-(x+y+z) = 270-240\\ &\Rightarrow\ y = 30\\ &\Rightarrow\ x+z = 240-30 = 210 \quad\text{(from (1))} \end{aligned}
4

Minimise xx under x≥zx\ge z. With x+z=210x+z=210 and x≥zx\ge z, the smallest xx occurs when x=zx=z.

x=z=2102=105\begin{aligned} &x = z = \frac{210}{2} = 105 \end{aligned}
xmin⁡=105x_{\min} = 105
CAT 2018 Slot 2 QA Q3: The arithmetic mean of x, y and z is 80, and that of x, y, z, u and v is 75, where u = (x + y)/2 and v = (y + — Solution | TheCATExam